Problem 60 - Calculus Early Transcendentals

Question

Verify that the conclusion of Clairaut's Theorem holds, that is, \(u_{x y}=u_{y x}\) $$u=x y e^{y}$$

Step-by-Step Solution

Answer

Yes, the conclusion of Clairaut's Theorem holds since \( u_{xy} = u_{yx} \).

Step 1: Differentiate with Respect to x

To determine if Clairaut's Theorem holds, first find the partial derivative of the function with respect to x. The function is given as \( u = x y e^y \). Since \( y \) and \( e^y \) are constants with respect to \( x \), differentiate as usual: \[ u_x = \frac{\partial}{\partial x}(x y e^y) = y e^y \] This is the partial derivative of \( u \) with respect to \( x \).

Step 2: Differentiate u_x with Respect to y

Now take the partial derivative of \( u_x = y e^y \) with respect to \( y \). Both \( y \) and \( e^y \) depend on \( y \), so use the product rule for derivatives: \[ u_{xy} = \frac{\partial}{\partial y}(y e^y) = (1)(e^y) + y(e^y) = e^y + y e^y = (1+y)e^y \] So, \( u_{xy} \) is \( (1+y)e^y \).

Step 3: Differentiate with Respect to y

Next, find the partial derivative of the original function with respect to \( y \). The function given is \( u = x y e^y \). Here, use the product rule for the factors \( y \) and \( e^y \) since both depend on \( y \): \[ u_y = \frac{\partial}{\partial y}(x y e^y) = x \left( y(e^y) + e^y \right) = x(1+y)e^y \] This is the partial derivative of \( u \) with respect to \( y \).

Step 4: Differentiate u_y with Respect to x

Finally, take the partial derivative of \( u_y = x(1+y)e^y \) with respect to \( x \). Since we treat everything except \( x \) as a constant, we have:\[ u_{yx} = \frac{\partial}{\partial x}( x(1+y)e^y ) = (1+y)e^y \] So, \( u_{yx} \) is \( (1+y)e^y \).

Step 5: Compare Mixed Partial Derivatives

Now we compare the mixed partial derivatives found: \[ u_{xy} = (1+y)e^y \] \[ u_{yx} = (1+y)e^y \] Both mixed partial derivatives are equal, thus verifying that \( u_{xy} = u_{yx} \), confirming the conclusion of Clairaut's Theorem.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus used to study functions with multiple variables. Unlike ordinary derivatives which deal with a function of a single variable, partial derivatives focus on how a function changes as one of its variables is altered, keeping the others constant. For instance, if we have a function \( u = x y e^y \), to find the partial derivative with respect to \( x \), treat \( y \) and \( e^y \) as constants. By differentiating with respect to \( x \), we get \( u_x = y e^y \). Similarly, when differentiating the function with respect to \( y \), apply the partial derivative rules to get the distinct derivative \( u_y \), which would consider \( x \) as a constant. This approach is critical in understanding how changes in one variable impact the function independently of others.

Partial derivatives are vital in fields that involve modeling and analyzing multivariable systems, such as physics, engineering, and economics. They form the basis for operations like gradient and Hessian computations, essential in optimization problems.

Product Rule for Derivatives

The product rule is a crucial tool in calculus that simplifies finding derivatives of functions composed of multiple factors. When dealing with partial derivatives, it's important to understand that the product rule applies similarly. For instance, when differentiating a function like \( u = x y e^y \) with respect to \( y \), use the product rule to tackle the factor \( y e^y \).

Apply the product rule:

• Differentiate the first function \( y \) and multiply by the second function \( e^y \).
• Then, differentiate \( e^y \) and multiply by \( y \).

This results in the expression \( (1)e^y + y(e^y) = (1 + y)e^y \).

In a broader sense, the product rule ensures accurate derivatives when products of algebraic, exponential, or trigonometric expressions arise. Understanding and utilizing the product rule effectively is crucial, especially in multivariable calculus, where functions can become more complex and interdependent.

Mixed Partial Derivatives

Mixed partial derivatives occur when we differentiate a multivariable function with respect to different variables in succession. In most situations, particularly for well-behaved functions, these mixed derivatives should be equal regardless of the order of differentiation. For instance, if we have already found \( u_{xy} \), then \( u_{yx} \) should equal \( u_{xy} \) for the theorem to hold.

These derivatives can be expressed as:

• \( u_{xy} \) - first differentiate with respect to \( x \), then \( y \).
• \( u_{yx} \) - first differentiate with respect to \( y \), then \( x \).

Upon computation using the original function \( u = x y e^y \), you'll observe both processes yield \( (1+y)e^y \). This equality confirms the conclusion of Clairaut's Theorem, reflecting the symmetry in the behavior of mixed derivatives, assuming continuity and smoothness of the function.

Mathematical Proof

Mathematical proofs are crucial in establishing the validity of mathematical statements or theorems. In the context of Clairaut's Theorem, the proof involves demonstrating the equality of mixed partial derivatives \( u_{xy} \) and \( u_{yx} \). A complete proof should not only calculate the derivatives but also consider the conditions under which the theorem holds, such as the continuity and smoothness of the function involved.

The basic steps in proving for \( u = x y e^y \) involve:

• Calculating \( u_x \) and \( u_y \) separately, accounting for how each variable independently affects the function.
• Using the product rule where applicable to compute accurate derivatives.
• Ensuring the continuity of the second partial derivatives, a necessary condition for Clairaut's Theorem.

By following these logical steps and comparing our results, the proof draws a conclusion reaffirming the symmetry in mixed partial derivatives, solidifying our understanding of multivariable calculus. This solidifies confidence in the mathematical framework used to analyze and interpret complex functions.